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Let α and β be the zeroes of the required polynomial f(x). 

Then (α + β) = \(\frac{5}2\) and αβ = 1 

∴ f(x) = x² - (α + β) x + αβ 

⇒ f(x) = x² - \(\frac{5}2\) x + 1 

⇒ f(x) = 2x² – 5x + 2 

Hence, the required polynomial is f(x) = 2x² – 5x + 2

∴ f(x) = 0 ⇒ 2x² – 5x + 2 = 0 

⇒ 2x² – (4x + x) + 2 = 0 

⇒ 2x² – 4x – x + 2 = 0 

⇒ 2x (x – 2) – 1 (x – 2) = 0 

⇒ (2x – 1) (x – 2) = 0 

⇒ (2x – 1) = 0 or (x – 2) = 0

⇒ x = \(\frac{1}2\) or x = 2 

So, the zeros of f(x) are \(\frac{1}2\) and 2.

since two  zeroes of the given polynomial  are 2 + √2 and 2 - √2 , the polynomial should have two factors (x-2-√2) and (x-2+√2) that means it has quadratic factor

(x-2-√2) X (x-2+√2)=(x-2)^2 -2= x²-4x+2

Now x⁴ − 8x³ + 19x² − 12x + 2

= x⁴-4x³+2x²-4x³+16x²-8x +x²-4x+2

=x²( x²-4x+2) − 4x ( x²-4x+2)+ ( x²-4x+2)

=( x²-4x+2)( x²-4x+1)

Hence other quadratic factor should be= (x4 − 8x3 + 19x2 − 12x + 20)/ ( x^2-4x+2)

=( x²-4x+1)
Solve ( x²-4x+1)=0

=> (x-2)²-3=0

hence other two zeroes are 2 + √3 and 2 - √3

We can find the quadratic equation if we know the sum of the roots and product of the roots by using the formula 

x² – (Sum of the roots)x + Product of roots = 0 

⇒ x² – √2x + \(\frac{1}3\) = 0 

⇒ 3x² –3√2x + 1 = 0

We know the sum, sum of the product of the zeroes taken two at a time and the product of the zeroes of a cubic polynomial then the cubic polynomial can be found as x³ – (sum of the zeroes)x² + (sum of the product of the zeroes taking two at a time)x – product of zeroes 

Therefore, the required polynomial is

x³ - 5x² -2x +24

If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as 

x³ – (a + b + c)x² + (ab + bc + ca)x – abc ……(1) 

Let a = 1/2 , b = 1 and c = –3 

Substituting the values in (1), we get 

x³ – ( 1/2 + 1 − 3)x² + ( 1/2 − 3 − 3/2 )x – ( −3/2 ) 

⇒ x³ – ( −3/2 )x² – 4x + 3/2 

⇒ 2x³ +3x² – 8x + 3

(c)   \(\frac{-3}{\sqrt2}\), \(\frac{\sqrt2}4\)

Let f(x) = 4x² + 5√2x – 3 = 0 

⇒ 4x² + 6√2x – √2x – 3 = 0 

⇒ 2√2x(√2x + 3) –1 (√2x + 3) = 0 

⇒ (√2x + 3) (2√2x – 1) = 0

⇒ x = – \(\frac{3}{\sqrt2}\) or x = \(\frac{1}{2\sqrt2}\)

⇒ x = – \(\frac{3}{\sqrt2}\)or x = \(\frac{1}{2\sqrt2}\) x \(\frac{\sqrt2}{\sqrt2}\) = \(\frac{\sqrt2}4\)

If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as 

x³ – (a + b + c)x² + (ab + bc + ca)x – abc ……(1) 

Let a = 2, b = –3 and c = 4 

Substituting the values in 1, we get 

x³ – (2 – 3 + 4)x² + (– 6 – 12 + 8)x – (–24) 

⇒ x³ – 3x² – 10x + 24

General form of a cubic polynomial whose zeroes are a, b and c is:

x³ – (a + b + c) x² + (ab + bc + ca)x – abc …(1)

To Find: Cubic polynomial whose zeroes are 2, -3 and 4

Let us say, a = 2, b = – 3 and c = 4

Putting the values of a, b and c in the equation (1) we get:

= x³ – (2 – 3 + 4) x² + (-6 – 12 + 8) x – (- 24)

= x³ – 3x² – 10x + 24

Which is required polynomial.

Let f(x) = 3x³ – 10x² – 27x + 10

5, -2 and 1/3 are the zeros of the polynomial (given)

Therefore,

f(5) = 3(5)³ – 10(5)² – 27(5) + 10

= 3 × 125 – 250 – 135 + 10

= 0

f(-2) = 3(-2)³ – 10(-2)² – 27(-2) + 10

= – 24 – 40 + 54 + 10

= 0

f(1/3) = 3 (1/3)³ – 10(1/3)² – 27(1/3) + 10

= 1/8 – 10/9 – 9 + 10

= 0

Verify relations:

General form of a cubic equation: ax³ + bx² + cx + d.

Now,

Consider α = 5, β = – 2 and γ = 1/3

α + β + γ = 5 – 2 + 1/3 = 10/3 = -b/a

αβ + βγ + αγ = 5 (-2) + (-2) (1/3) + (1/3) (5) = -9 = c/a

And, αβγ = 5 (-2) (1/3)= -10/3 = -d/a

The value of s(z) at z = 1 is 1³ – 1 = 0

x² + 1 = 0 

⇒ x² = -1 

No real number exists such that whose root is – 1. 

∴ x² + 1 has no zeroes.

(99)³ = (100 – 1)³ 

= 100³ – 3 (100)² (1) + 3 (100) (1)² – 1³ 

[ ∵ (x – y)³ = x³ – 3x²y + 3xy² + y³ ] 

= 1000000 – 30000 + 300 – 1 

= 970299

Given: (x + a) is a factor of 2x² + 2ax + 5x + 10 

We have 

x + a = 0 

⇒ x = –a 

Since, (x + a) is a factor of 2x² + 2ax + 5x + 10 

Hence, It will satisfy the above polynomial 

∴ 2(–a)² + 2a(–a) + 5(–a) + 10 = 0 

⇒ –5a + 10 = 0 

⇒ a = 2

Equating x² – x to 0 to find the zeroes, we will get 

x(x – 1) = 0 

⇒ x = 0 or x – 1 = 0 

⇒ x = 0 or x = 1 

Since, x³ + x² – ax + b is divisible by x² – x. 

Hence, the zeroes of x² – x will satisfy x³ + x² – ax + b 

∴ (0)³ + 0² – a(0) + b = 0 

⇒ b = 0 

And (1)³ + 1² – a(1) + 0 = 0 [∵b = 0] 

⇒ a = 2 

Let f(x) = 5x² ˗ 4 ˗ 8x

= 5x² ˗ 8x ˗ 4

= 5x² ˗ (10x ˗ 2x) ˗ 4

= 5x² ˗ 10x + 2x ˗ 4

= 5x (x ˗ 2) + 2(x ˗ 2)

= (5x + 2) (x ˗ 2)

To find the zeroes, set f(x) = 0

(5x + 2) (x ˗ 2) = 0

5x + 2 = 0 or x ˗ 2 = 0

x = (−2)/5 or x = 2

Again,

Sum of zeroes = (-2)/5 + 2 = (-2+10)/5 = 8/5

= -b/a

= (-Coefficient of x)/(Cofficient of x²)

Product of zeroes = (-2/5) x 2 = (-4)/5

= c/a

= Constant term / Coefficient of x²

Let f(x) = 2 √3 x² – 5x + √3

= 2 √3 x² – 2x – 3x + √3

= 2x(√3x-1) – √3(√3x-1)

To find the zeroes, set f(x) = 0

(√3x – 1) or (2x – √3) = 0

x = 1/√3 = √3/3 or x = √3/2

x = √3/3 or x = √3/2

Again,

Sum of zeroes = √3/3 + √3/2 = 5√3/6

= -b/a

= (-Coefficient of x)/(Cofficient of x²)

Product of zeroes = √3/3 x √3/2 = √3/6

= c/a

= Constant term / Coefficient of x²

Let the zeroes of the quadratic polynomial be

α = 3 , β = – 3

Then, α + β = 3 + ( – 3) = 0

αβ = 3 × ( – 3) = – 9

Sum of zeroes = α + β = 0

Product of zeroes = αβ = – 9

Then, the quadratic polynomial

= x² – (sum of zeroes)x + product of zeroes

= x² – (0)x + ( – 9)

= x² – 9

Let f(x) = 2x² ˗ 11x + 15

= 2x² ˗ (6x + 5x) + 15

= 2x² ˗ 6x ˗ 5x + 15

= 2x (x ˗ 3) ˗ 5 (x ˗ 3)

= (2x ˗ 5) (x ˗ 3)

To find the zeroes, set f(x) = 0

(2x ˗ 5) (x ˗ 3) = 0

2x ˗ 5= 0 or x ˗ 3 = 0

x = 5/2 or x = 3

Again,

Sum of zeroes = 5/2 + 3 = (5+6)/2 = 11/2

= -b/a

= (-Coefficient of x)/(Cofficient of x²)

Product of zeroes = 5/2 x 3 = 15/2

= c/a

= Constant term / Coefficient of x²

Let f(x) = 4x² ˗ 4x + 1

= (2x²) – 2(2x)(1) + (1)^2

= (2x – 1)²

To find the zeroes, set f(x) = 0

(2x – 1)² = 0

x = 1/2 or x = 1/2

Again,

Sum of zeroes = 1/2+1/2=1=1/1

= -b/a

= (-Coefficient of x) / (Cofficient of x²)

Product of zeroes = 1/2 x 1/2=1/4

= c/a

= Constant term / Coefficient of x²

Given: Sum of zeroes = α + β = 4

Product of zeroes = αβ = 1

Then, the quadratic polynomial

= x² – (sum of zeroes)x + product of zeroes

= x² – (4)x + 1

= x² – 4x + 1

Product of zeroes = αβ = 1

Then, the quadratic polynomial

= x² – (sum of zeroes)x + product of zeroes

= x² – (1)x + 1

= x² – x + 1

Given: Sum of zeroes = α + β = 0

Product of zeroes = αβ = 3

Then, the quadratic polynomial

= x² – (sum of zeroes)x + product of zeroes

= x² – (0)x + 3

= x² + 3

The correct option is: (B) (2x² + 3)L

Explanation:

Capacity of both the containers is

 (2x³ + 2x² + 3x + 3) L and (4x³ - 2x² + 6x - 3)L 

i.e.,  (2x² + 3)(x + 1)L and (2x² + 3)(2x - 1)L 

Required measure is the H.C.F. of capacity of both the containers i.e., (2x² + 3) L

The correct option is: (A) Rs(2x³ - 18x² - 15)

Explanation: 

Total amount Raghav had = Rs(6x³ +2x² + 3x)

Cost of one shirt = Rs(x + 5)

Number of shirts he bought = 4x² + 3

. ^.. Amount spent by him = Rs(x + 5)(4x² + 3)

= Rs(4x³ + 20x² + 3x + 15)

Hence, money left with Raghav

= Rs(6x³ + 2x² + 3x - 4x³ - 20x² - 3x - 15)

= Rs(2x³ - 18x² - 15)

The correct option is: (B) (-6x² + 9) m

Explanation:

Length of the garden = (2x³ - 7x² + 7) m

Perimeter of the garden = 2 x (length + breadth)

. ^.. 4x³ - 2x² + 4 = 2(2x³ + 5x² - 7 + breadth)

=> 2x³ + 5x² + 2 = (2x³ + 5x² - 7) + breadth

So, breadth ofthe rectangle

2x³ + 5x² + 2 - 2x³ + 5x² - 7 = (-6x² + 9) m

No, x⁴ - 1 is a polynomial intersecting the x-axis at exactly two points.

If the graph of a polynomial intersects the x - axis at exactly two points, it need not be a quadratic polynomial: True.

The polynomial with degree greater than 2 can also intersect the x - axis at two points. Hence, it need not be a quadratic polynomial.

No, since degree (x – 1) = 1 = degree (2x + 3).

1. Parabola

2. c) a < 0

3. c) 2

4. b) -2, 4

5. b) \(-\cfrac2{\sqrt3},\cfrac{\sqrt3}4\)

Given: x = –2 is one zero of the polynomial 3x² + 4x + 2k 

Therefore, it will satisfy the above polynomial. 

Now, we have 

3(–2)² + 4(–2)1 + 2k = 0 

⇒ 12 – 8 + 2k = 0 

⇒ k = – 2 

Given: x = 1 is one zero of the polynomial ax² – 3(a – 1) x – 1 

Therefore, it will satisfy the above polynomial. 

Now, we have 

a(1)² – (a – 1)1 – 1 = 0 

⇒ a – 3a + 3 – 1 = 0 

⇒ –2a = – 2 

⇒ a = 1 

Answer is (C) 4x³y²

Let the side of a square is x m.

Perimeter of that square = 4x m

Difference in perimeter is 24 m

Perimeter of second square 

= 4x + 24 m

Then, side of second square 

= (4x + 24)/4 = 4(x + 6)/4 = (x + 6)m

Area of first square = x² sq m

Area of second square 

= (x + 6)² sq m = x² + 12x + 36 sq m

Sum of areas of both squares = 468 sq m

x² + (x² + 12x + 36) = 468

⇒ 2x² + 12x + 36 – 468 = 0

⇒ 2x²+ 12x – 432 =0

⇒ 2(x² + 6x – 216) = 0

⇒ x² + 6x – 216 = 0

⇒ x² + 18x – 12x – 216 = 0

⇒ x(x + 18) – 12(x + 18) = 0

⇒ (x + 18)(x – 12) = 0

when x + 18 = 0, then x = -18 (not possible)

or x – 12 = 0, then x = 12

∴ x = 12

Side of smaller square = 12 m

and side of larger square 

= x + 6 = 12 + 6 

= 18 m

Thus sides of both  squares are 12 m. and 18 m respatively

Correct option is (B) -1

x+2 is a factor of \(2x^4+3x^3+2Kx^2+3x+6\)

\(\therefore\) x = -2 is a root of equation \(2x^4+3x^3+2Kx^2+3x+6=0\)

\(\therefore2(-2)^4+3(-2)^3+2k(-2)^2+3(-2)+6=0\)

\(\Rightarrow\) 32 - 24 + 8k - 6 + 6 = 0

\(\Rightarrow\) 8k = -32+24 = -8

\(\Rightarrow\) \(k=\frac{-8}8=-1\)

Correct option is B) -1

Answer is (D) None of these

Let p(y) = 3y³ + 8y² – 1

p(1) = 3 × (1)³ + 8(1)² – 1 

= 3 + 8 – 1 = 10 ≠ 0

p(-1) = 3(-1)³ + 8(-1)² – 1 

= -3 + 8 – 1 = 4 ≠ 0

and p(0) = 3 × (0)³ + 8(0)² – 1 

= 0 + 0 – 3 = -3 ≠ 0

Thus by putting y = 1, -1 the expression obtained is not equal to zero.

p(x) = (x –2)²−(x + 2)²

We know that,

Zero of the polynomial p(x) = 0

Hence, we get,

⇒ (x–2)²−(x + 2)² = 0

Expanding using the identity, a² – b² = (a – b) (a + b)

⇒ (x – 2 + x + 2) (x – 2 –x – 2) = 0

⇒ 2x ( – 4) = 0

⇒ – 8 x= 0

Therefore, the zero of the polynomial = 0

If all three zeroes of a cubic polynomial x³ + qx² - bx + c are positive, then at least one of a, b and c is non - negative: True.

Let α, β and γ be the zeroes of the polynomial p(x) = ax³ + bx² + cx + d, where α, β, γ > 0

Product of all the zeroes = - (constant term) ÷ coefficient of x³

αβγ = - d/a > 0 (∵ α, β, γ > 00 ⇒ αβγ > 0)

⇒ d/a < 0

⇒ d and a have different signs.

Sum of the products of two zeroes at a time = coefficient of x ÷ coefficient of x³

αβ + βγ + αγ = c/a > 0 (∵ α, β, γ > 0 ⇒ αβ, βγ, αγ > 0 ⇒ αβ + βγ + αγ > 0 )

⇒ c and a have the same signs.

Sum of the zeroes = - (coefficient of x²) ÷ coefficient of x³

α + β + γ = - b/a > 0 (∵ α, β, γ > 0 ⇒ α + β + γ > 0)

⇒ b/a < 0

⇒ b and a have different signs.

Case1: when a > 0 

c > 0 , b < 0 and d < 0

Case2: when a < 0 

c < 0 , b > 0 and d > 0

∴ In both cases two of the coefficients are non - negative.

In figure, the polynomial has 2 zeroes because the graph cuts x - axis at 2 points

i.e. x = – 3 and x = – 1.

If all the zeroes of a cubic polynomial are negative, then all the coefficients and the constant term of the polynomial have the same sign: True

Let α, β and γ be the zeroes of the polynomial p(x) = ax³ + bx² + cx + d, where α, β, γ < 0

Product of all zeroes = - (constant term) ÷ coefficient of x³

αβγ = - d/a < 0 (∵ α, β, γ < 0 ⇒ αβγ < 0)

⇒ d/a > 0

⇒ d and a have the same signs.

Sum of the products of two zeroes at a time = coefficient of x ÷ coefficient of x³

αβ + βγ + αγ = c/a > 0 (∵ α, β, γ < 0 ⇒ αβ,βγ,αγ > 0 ⇒ αβ + βγ + αγ > 0 )

⇒ c and a have the same signs.

Sum of the zeroes = - (coefficient of x²) ÷ coefficient of x³

α + β + γ = - b/a < 0 (∵ α, β, γ < 0 ⇒ α + β + γ < 0)

⇒ b/a > 0

⇒ b and a have same signs.

⇒ a,b,c and d have same signs.

In the figure,

The polynomial has 4 real zeroes because the graph cuts x - axis at four points.

b ² – 4ac = 0,

Two

The given quadratic equation touches the x - axis at only one point.

The root of the quadratic equation is equal and real because if the quadratic equation has two distinct roots, then the graph touches the x - axis at two points.

As we know that the roots are real and equal if the value of discriminant is zero,

So, b ² – 4ac = 0

In figure, the polynomial has 3 real zeroes because the graph cuts axis at three points.

The y - intercept of the equation is c,

So,

The signs of will be negative as c < 0

As seen from the graph,

The parabola cuts the graph at two points on the x axis.

One root is positive & one root is negative.

Now, for a polynomial, the sum of roots is given as:

α + β = -b/a

Also, the product of roots = c/a is positive.

Because c is positive,

⇒ a is negative & b is negative.

Therefore, option (b) is correct.

The y - intercept of the equation is c,

So,

The signs of will be positive as c > 0

The signs of a will be positive as a > 0 and,

The signs of b² – 4ac will be positive as b² – 4ac > 0

As seen from the graph,

The parabola cuts the graph at two points on the positive x - axis.

Hence, both the roots are positive.

Now, for a polynomial, the sum of roots is given as:

α + β = - b/a

∴ the sum will be positive as the roots are positive.

Also, the product of roots = c/a has to be positive too.

⇒ a is positive.

Now, since a is positive, therefore for the sum of roots to be negative, b has to be negative.

⇒ a > 0, b< 0 & c > 0.

Therefore, option (a) is correct.